Simplify the following expression and state the condition under which the simplification is valid. $y = \dfrac{k^3 + 10k^2 + 24k}{9k^2 - 27k - 486}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ y = \dfrac {k(k^2 + 10k + 24)} {9(k^2 - 3k - 54)} $ $ y = \dfrac{k}{9} \cdot \dfrac{k^2 + 10k + 24}{k^2 - 3k - 54} $ Next factor the numerator and denominator. $ y = \dfrac{k}{9} \cdot \dfrac{(k + 6)(k + 4)}{(k + 6)(k - 9)}$ Assuming $k \neq -6$ , we can cancel the $k + 6$ $ y = \dfrac{k}{9} \cdot \dfrac{k + 4}{k - 9}$ Therefore: $ y = \dfrac{ k(k + 4)}{ 9(k - 9)}$, $k \neq -6$